Friction 2 Worksheet - Free Printable

Let's solve each problem step by step from the Friction 2 Worksheet. We'll use physics principles related to friction, Newton’s Laws, and forces.

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Problem 1:

> A horizontal force of 300.0 N is required to pull a 145.0 kg trunk across the floor at a constant speed. Find the coefficient of kinetic friction.

#### Given:
- Applied force, $ F_{\text{app}} = 300.0 \, \text{N} $
- Mass, $ m = 145.0 \, \text{kg} $
- Constant speed → acceleration = 0 → net force = 0

#### Step 1: Understand motion
Since the trunk moves at constant speed, the net force is zero:
$$
F_{\text{net}} = F_{\text{app}} - f_k = 0 \Rightarrow f_k = F_{\text{app}} = 300.0 \, \text{N}
$$

#### Step 2: Use kinetic friction formula
$$
f_k = \mu_k \cdot N
$$
Normal force $ N $ on a horizontal surface (no vertical acceleration):
$$
N = mg = 145.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 1421 \, \text{N}
$$

Now solve for $ \mu_k $:
$$
\mu_k = \frac{f_k}{N} = \frac{300.0}{1421} \approx 0.211
$$

✔ Answer: $ \boxed{0.211} $

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Problem 2:

> How much force must be applied to push a 1.25 kg book across the desk at a constant speed if the coefficient of kinetic friction is 0.30?

#### Given:
- Mass, $ m = 1.25 \, \text{kg} $
- $ \mu_k = 0.30 $
- Constant speed → net force = 0 → applied force = friction force

#### Step 1: Find normal force
$$
N = mg = 1.25 \times 9.8 = 12.25 \, \text{N}
$$

#### Step 2: Find kinetic friction
$$
f_k = \mu_k \cdot N = 0.30 \times 12.25 = 3.675 \, \text{N}
$$

Since the book moves at constant speed, applied force must equal friction:
$$
F_{\text{app}} = f_k = 3.675 \, \text{N}
$$

✔ Answer: $ \boxed{3.68 \, \text{N}} $ (rounded to two decimal places)

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Problem 3:

> A force of 115.0 N is applied horizontally to a 25.0 kg box to move it across a horizontal floor. If the box has an acceleration of 3.00 m/s², find the coefficient of friction.

#### Given:
- $ F_{\text{app}} = 115.0 \, \text{N} $
- $ m = 25.0 \, \text{kg} $
- $ a = 3.00 \, \text{m/s}^2 $
- Find $ \mu_k $

#### Step 1: Use Newton’s Second Law
Net force:
$$
F_{\text{net}} = ma = 25.0 \times 3.00 = 75.0 \, \text{N}
$$

But:
$$
F_{\text{net}} = F_{\text{app}} - f_k
\Rightarrow f_k = F_{\text{app}} - F_{\text{net}} = 115.0 - 75.0 = 40.0 \, \text{N}
$$

#### Step 2: Normal force
$$
N = mg = 25.0 \times 9.8 = 245 \, \text{N}
$$

#### Step 3: Solve for $ \mu_k $
$$
\mu_k = \frac{f_k}{N} = \frac{40.0}{245} \approx 0.163
$$

✔ Answer: $ \boxed{0.163} $

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Problem 4:

> A 1400.0 N force is applied on a 250.0 kg crate to move it across the floor. The coefficient of friction between the crate and the floor is 0.22.

#### a) What is the net force on the crate?

#### Given:
- $ F_{\text{app}} = 1400.0 \, \text{N} $
- $ m = 250.0 \, \text{kg} $
- $ \mu_k = 0.22 $

#### Step 1: Normal force
$$
N = mg = 250.0 \times 9.8 = 2450 \, \text{N}
$$

#### Step 2: Friction force
$$
f_k = \mu_k \cdot N = 0.22 \times 2450 = 539 \, \text{N}
$$

#### Step 3: Net force
$$
F_{\text{net}} = F_{\text{app}} - f_k = 1400.0 - 539 = 861 \, \text{N}
$$

✔ Answer (a): $ \boxed{861 \, \text{N}} $

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#### b) What is the acceleration of the crate?

Use Newton’s second law:
$$
a = \frac{F_{\text{net}}}{m} = \frac{861}{250.0} = 3.444 \, \text{m/s}^2
$$

✔ Answer (b): $ \boxed{3.44 \, \text{m/s}^2} $ (rounded to two decimal places)

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✔ Final Answers Summary:

1. $ \boxed{0.211} $
2. $ \boxed{3.68 \, \text{N}} $
3. $ \boxed{0.163} $
4.
a. $ \boxed{861 \, \text{N}} $
b. $ \boxed{3.44 \, \text{m/s}^2} $

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